Chapter 3.2, Problem 79E

(a)

To determine

To sketch: the graph of the function $P\left(x\right)=\left(x-2\right)\left(x-4\right)\left(x-5\right)$ and to find the number of local extrema.

Answer to Problem 79E

The total number of extrema is 2.

Explanation of Solution

Given information:

  $P\left(x\right)=\left(x-2\right)\left(x-4\right)\left(x-5\right)$

Calculation:

Consider the function,

  $P\left(x\right)=\left(x-2\right)\left(x-4\right)\left(x-5\right)$

Rewrite the function as,

  $\begin{array}{c}P\left(x\right)=\left(x-2\right)\left(x-4\right)\left(x-5\right)\\ =\left(x-2\right)\left(x^2-9x+20\right)\\ =x^3-9x^2+20x-2x^2+18x-40\\ =x^3-11x^2+38x-40\end{array}$

The graph of $P\left(x\right)=x^3-11x^2+38x-40$ is shown below:

  

From the graph, it is observed that there are two local extrema for the function,

  $P\left(x\right)=\left(x-2\right)\left(x-4\right)\left(x-5\right)$

Thus, the total number of extrema is 2.

(b)

To determine

To explain: the reason of the given function must have two local extrema.

Answer to Problem 79E

There are two local extrema for the function $P\left(x\right)=\left(x-a\right)\left(x-b\right)\left(x-c\right)$ when $a<b<c$ .

Explanation of Solution

Given information:

Given function,

  $P\left(x\right)=\left(x-a\right)\left(x-b\right)\left(x-c\right)$

If $a<b<c$

Calculation:

If $P$ is a polynomial function and $P\left(a\right)$ and $P\left(b\right)$ have opposite signs, then there exist one value of $c$ in between $a$ and $b$ such that $P^{\prime }\left(c\right)=0$

The graph of the function $P\left(x\right)=\left(x-a\right)\left(x-b\right)\left(x-c\right)$ when $a<b<c$ is shown below:

  

Observe the graph, it is clear that the function is increasing on the interval $\left(a,b\right)$ and decreasing on the interval $\left(b,c\right)$ and the derivative is positive before $b$ and negative on $\left(b,c\right)$ . Since $P^{\prime }\left(x\right)>0$ on $\left(a,b\right)$ and $P^{\prime }\left(x\right)<0$ on $\left(b,c\right)$ , then by the intermediate value theorem,

  $P^{\prime }\left(x\right)=0$ for some $x\in \left(a,b\right)$

In fact, $P^{\prime }\left(x\right)=0$ at $x=b$

This means that the function $P\left(x\right)$ has a relative maximum at $x=b$

In the same situation, in reverse, happens at $x=c$ , where $P\left(x\right)$ has a relative minimum, Thus there are two local extrema for the function $P\left(x\right)=\left(x-a\right)\left(x-b\right)\left(x-c\right)$ when $a<b<c$ .